Question

The expression $[x+(x^3-1)^{1/2}]^5 +[x-(x^3-1)^{1/2}]^5$ is a polynomial of degree

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

Answer: (C)

We know that,
$(a+b)^5+(a-b)^5= \,{}^5C_0 a^5+ \,{}^5C_1 a^4b+ \,{}^5C_2 a^3b^2$
$ + ^5C_3a^2b^3+ \,{}^5C_4ab^4+ \,{}^5C_5b^5+^5C_0 a^5- \,{}^5C_1 a^4b$
$ + ^5 C_2 a^3b^2- \,{}^5C_4 ab^4+ \,{}^5C_3 a^2b^3+ \,{}^5C_4 ab^4- \,{}^5C_5 b^5$
$ =2[a^5+10a^3+b^2+5ab^4]$
$\therefore \, [x+(x^3-1)^{1/5}]^5 + [ x-(x^3-1)^{1/2}]^5$
=2 $[x^5+10x^3(x^3-1)+5x(x^3-1)^2]$
Therefore, the given expression is a polynomial of degree $7$.